Question

considere a PG (5/3, -10/9, 20/27, -40/81, ...) o oitavo termo dessa sequência é?

Answer 1

Olá,

na P.G. acima, temos que:

$\begin{cases}a_1= \dfrac{5}{3}\\ q= \dfrac{a_2}{a_1}~\to~q= -\dfrac{10}{9}\div \dfrac{5}{3}~\to~q=- \dfrac{10}{9}\cdot \dfrac{3}{5}~\to~q=- \dfrac{2}{3}\\ n=8~termos\\ a_8=? \end{cases}$

Inserindo esses dados, na fórmula do termo geral da P.G.:

$a_n=a_1\cdot q^{n-1}\\\\ a_8= \dfrac{5}{3}\cdot\left( -\dfrac{2}{3}\right)^{8-1}\\\\ a_8= \dfrac{5}{3}\cdot\left(- \dfrac{2}{3}\right)^7\\\\ a_8= \dfrac{5}{3}\cdot\left(- \dfrac{128}{2.187}\right)\\\\\\ \Large\boxed{\boxed{a_8=- \dfrac{640}{6.561}}}.\\.$

Tenha ótimos estudos, FLW!